## College Algebra (11th Edition)

$x=\left\{ 3,5 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given rational equation, $\dfrac{2x-5}{x}=\dfrac{x-2}{3} ,$ use cross-multiplication and the concepts of quadratic equations. $\bf{\text{Solution Details:}}$ Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} (2x-5)(3)=x(x-2) .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac$ and the properties of equality, the expression above is equivalent to \begin{array}{l}\require{cancel} 6x-15=x^2-2x \\\\ -x^2+(6x+2x)-15=0 \\\\ -x^2+8x-15=0 \\\\ -1(-x^2+8x-15)=(0)(-1) \\\\ x^2-8x+15=0 .\end{array} In the trinomial expression above the value of $ac$ is $1(15)=15$ and the value of $b$ is $-8 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -3,-5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2-3x-5x+15=0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^2-3x)-(5x-15)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x-3)-5(x-3)=0 .\end{array} Factoring the $GCF= (3x-4)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x-3)(x-5)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-3=0 \\\\\text{OR}\\\\ x-5=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-3=0 \\\\ x=3 \\\\\text{OR}\\\\ x-5=0 \\\\ x=5 .\end{array} Upon checking, none of the solutions above will make the denominator of the original equation equal to zero. Hence, $x=\left\{ 3,5 \right\} .$